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3 years ago

Please help! I need answers in begging !

Mathematics

1 answer:

Levart [38]3 years ago

7 0

Answer:

2866

Step-by-step explanation:

10000-7134=2866

so 7134+2866=10000

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Suppose 67% of all teenagers own a laptop and 29% of all teenagers own a laptop and a tablet. What is the probability that a tee

Darya [45]

The answer to your question is: B

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3 years ago

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Hello can anyone help me please

Naya [18.7K]

Answer:

5.3 to the 4th power :)

Step-by-step explanation:

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3 years ago

Jess deposited $5,000 into an account that earns simple interest. After 9 years, Jess had earned $3,150 in interest. What was th

musickatia [10]

Answer:

The interest rate of Jess's account was 7%

Step-by-step explanation:

A = P * (1 + rt)

A = final amount

P = initial principal balance

r = annual interest rate

t = time (in years)

Replacing with the values we know:

A = P * (1 + rt)

8,150 = 5,000 * (1 + r * 9)

8,150/5,000 = 1 + 9r

1.63 = 1 + 9r

1.63 - 1 = 9r

0.63 = 9r

r = 063/9

r = 0.07 = 7%

The interest rate of Jess's account was 7%

4 0

3 years ago

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What number does x stand for in this equation?

Alex777 [14]

Answer:

C. -3

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Step-by-step explanation:

Step 1: Define

-7x + 6 = 27

Step 2: Solve for x

[Subtraction Property of Equality] Subtract 6 on both sides: -7x = 21
[Division Property of Equality] Divide -7 on both sides: x = -3

3 0

2 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

8 0

3 years ago